Download Lecture 15

Electric Fields in Materials
insulators : do not have moving charges HdielectricsL,
charge is fixed
conductors : have moving charges
Σ = conductivity
Σ >> 1 conductor such as a metal
Σ << 1 insulator such as glass
Σ » 1 semiconductor such as silicon,
germanium
I=
dQ
dt
Ó
Ó Ó
; I = à J × d S, J = current density
Convection current : flow of charge through space
Convection current
DQ
ΡV DSDl
Ó Ó
DI =
=
= ΡV D S.u
Dt
Dt
Ó
Ó
J = ΡV u
2
lect15new.nb
DQ
Dt
=
ΡV DSDl
Dt
Ó Ó Ó Ó
= ΡV D S.u = J.D S
Conduction current : flow of charge in materials
electrons bump into lattice structure of the material which results
in the slowing of the charge
3
lect15new.nb
Ó
Dmomentum
Ó Ó mu
-e E = F =
=
Τ
time
drift velocity
Ó
Ó
Ó -e E Τ
u=
= -Μe E
m
Μe =
eΤ
m
= electron mobility
Τ = transit time of an electron, m = mass of an electron
Ó
eEΤ
Ó
Ó
Ó
J = ΡV u = -ΡV
= -ΡV Μe E
m
ΡV = -n e
2
Ó ne Τ Ó
Ó
Ó
J=
E = ΡV Μe E = Σ E
m
ne2 Τ
Σ = ΡV Μe =
= conductivity
m
Ó
Ó
Ó
For semiconductor, J = H-ΡVe Μe + ΡVh Μh L E = Σ E
Ó
1. Apply E field
2. Charge accumulates on the conductor surface
3. Charge moves inside the conductor so as to cancel the effect of
Ó
the applied E field
Ó
4. Charge stops, net result : internal E field Ei
Ó
5. Ei + E = 0
Ó
Net E field inside a perfect conductor is 0
Σ = ¥ for a perfect conductor
4
lect15new.nb
perfect conductor : V = constant
Ó
E = -Ñ V = 0
-Ñ V =
¶v ` ¶v ` ¶v `
x+
y+
z =0
¶x
¶y
¶z
V is constant for a perfect conductor
As temp increases, Σ decreases,
because there is more knocking into lattice
non - perfect conductor
Ó
Ó Ó
Ó
J = Σ E, E points in the same direction as J, for many resistors,
Ó
J is in one direction Hlong, thin resistorL
Ó
Ó
V = - à E × d l = E l,
E = Vl
5
lect15new.nb
Ó
Ó
I = à J × d S = J A, J = I  A
Ó
Ó
J = Σ E, J = Σ V  l = I  A
Ohm' s Law :
l
l
R = VI =
=Ρ
ΣA
A
1
Ρ=
= resistivity W - m
Σ
Σ = °  m siemans  m
Ó
Ó
Ó
Ó
-Ù E × d l
-Ù E × d l
R = VI =
=
Ó
Ó
Ó
Ó
ÙJ ×d S
ÙΣ E×d S
5.1
j = 81  r^3 ´ 2 Cos@ΘD, 1  r^3 Sin@ΘD, 0<
HaL
i = Integrate@1  r^3 ´ 2 Cos@ΘD r^2 Sin@ΘD, 8Θ, 0, Π<, 8Φ, 0, Π<D . r ® .2
0
i = Integrate@1  r^3 ´ 2 Cos@ΘD r^2 Sin@ΘD, 8Θ, 0, Π  2<, 8Φ, 0, 2 Π<D . r ® .2
31.4159
HbL
i = Integrate@1  r^3 ´ 2 Cos@ΘD r^2 Sin@ΘD, 8Θ, 0, Π<, 8Φ, 0, 2 Π<D . r ® .1
0
6
lect15new.nb
5.2 A typical example of convective charge transport is found in the
Van de Graaff generator where charge is transported on a moving
belt from the base to the dome as shown in Figure 5.4.If a surface
charge density Ρs = 10 - 7 C  m2 is transported at a velocity of u = 2 m  s,
calculate the charge collected in 5 s.Take the width of the belt as w =
10 cm.
I = Ρs u w; Q = I t = Ρs u w t = 100 nC
5.3
A wire of diameter 1 mm and conductivity 5 * 10^7 S  m has 10^29 free
electrons  m3 when an electric field of 10 mV  m is applied.Determine
HaL The charge density of free electrons
HbL The current density
HcL The current in the wire
HdL The drift velocity of the electrons.Take the electronic charge as
e = -1.6 * 10^-19 C.
d = .001; Σ = 5 ´ 10^7; n = 10^29; efield = .01;
e = -1.6 10^-19;
charge density :
Ρv = e n
10
-1.6 10
current density :
j = Σ efield
500000.
7
lect15new.nb
area = Hd  2L^2 Pi
-7
7.85398 10
current :
i = j area
0.392699
drift velocity :
Solve@j == Ρv u, uD
{{u -> -0.00003125}}
Ó
Ó
V = à E×d l = E l
Ó
Ó
negative sign is dropped because à E × d l < 0 if I > 0
J = Σ E, I = J A, J = I  A, I  A = Σ V  l
l
l
1
V=I
=I Ρ
, Ρ=
ΣA
A
Σ
8
lect15new.nb
Ó
Ó
Ó
Ó
ÙE×d l
ÙE×d l
R = VI =
=
Ó
Ó
Ó
Ó
ÙJ ×d S
ÙΣ E×d S
Ρ : W - m, R = W
Power : P =
dW
dt
Ó
Ó
Ó Ó
Ó Ó
Ó
Ó
Ó
Ó
W = à F × d l, P = à F × u = Kà à à O Ρv dv E × u, JF = Ρv dv EN, J = Ρv u
Ó Ó
P = à à à E × J dv
L
L
Power density : wp =
dv = dl dS
dP
dv
P = I2 R HJoule' s LawL
Ó Ó
Ó Ó
= E × J = Σ E2 , P = à à à E × J dv = à E dl à J dS = V I,
L
s
5.4 A lead Hs = 5 * 10^6 S  mL bar of square cross section has a hole
bored along its length of 4 m so that its cross section becomes that
of Figure below .Find the resistance between the square ends.
R=Ρ
l
A
, Ρ = 1  Σ, l = 4, A = .03^2 - Pi .005^2
Σ = 5 ´ 10^4;
A = area
9
lect15new.nb
area = .03^2 - Pi .005^2
0.00082146
l = 4;
Ρ = 1Σ
1
50 000
R=Ρ
l
area
0.0973876
units are W
Polarization : occurs in a dielectric HinsulatorL in the presence of an E field, formulation of dipoles of alignment of dipoles from E - field
Ρps = surface charge density due to polarization
Ρpv = volume charge density due to polarization
10
lect15new.nb
Ó Ú Qi di
P=
Dv
dV =
Ñ'
lim Dv®0
dv'
Ó `
P×R
4 Π¶o
R ^2
1
=
R
`
R
R2
;
Ó `
P×R
R ^2
Ó
= P×Ñ'
1
R
Ó
Ó
Ó
A×Ñ' f = Ñ'×f A-f Ñ'×A
Ó
P×Ñ'
V=à
à
1
R
dv'
4 Π¶o
1
4 Π¶o
= Ñ'×
Ñ'×
1 Ó 1
Ó
P- Ñ'×P
R
R
1 Ó 1
Ó
P- Ñ'×P =
R
R
1 Ó `
dv' 1
Ó
P × n dS - à
Ñ'×P
R
4 Π¶o R
Ó `
Ó
Ρps = P × n; Ρpv = -Ñ ' × P
à Ρps dS + à Ρpv dv = 0
Material was neutral before E - Field was applied so it is still neutral
11
lect15new.nb
Ó `
Ρps = P × n
Ó
Ρpv = -Ñ × P
Let' s consider a dielectric with free charge Ρv
Ó
ΡT = Ρv + Ρpv = Ñ × JΕo EN
Ó
Ó
Ρv = Ñ × JΕo EN - Ρpv = Ñ × D =
Ó
Ó
Ó Ó
Ñ × JΕo EN + Ñ × P = Ñ × JΕo E + PN,
Ó
Ó Ó
D = Εo E + P
Ó
Ó
P = Χ Εo E, Χ = electric susceptance,
Εo + Χ Εo = Ε,
Ε = permittivity
Εo = permittivity in free space,
Ε = Εr Εo , Εr = relative permittivity
dielectrics are used in capacitors
The greater Ε, the greater the energy stored
Example 5.5
12
lect15new.nb
A dielectric cube of side L and center at the origin has a radial
Ó
polarization given by P = a r,
Ó
where a is a constant and r =
`
`
`
x x + y y + z z.Find all bound charge densities and show explicitly
that the total bound charge vanishes.
dielectric cube of side L, centered at origin, has
Ó
P = a 8x, y, z<
Find Qs
Ó `
Ρps = P × n
For each side : Ρps = a L  2, Qside = a L3 ‘ 2 therefore, the total =
Qs = 3 a L3
Find Qv
Ó
Ρpv = -Ñ × P = -Ñ × Ha 8x, y, z<L = -3 a
Therefore Qv = -3 a L3
which equals - Qs , so that the total bound charge vanishes
Ó
Ó
Ó
Ó
ÙE×d l
ÙE×d l
R = VI =
=
Ó
Ó
Ó
Ó
ÙJ ×d S
ÙΣ E×d S
Ó
Ó
Ó
Ó
ÙD×d S
ÙΕ E×d S
C = QV =
=
Ó
Ó
Ó
Ó
ÙE×d l
ÙE×d l
13
lect15new.nb
Parallel plate capacitor
Ó
Ó
à E × d l = E d = V, E = V  d
Ó
Ó
à Ε E×d S = Ε E A
Ó
Ó
Ó
Ó
ÙD×d S
ÙΕ E×d S
ΕEA
Ε A
C = QV =
=
=
=
Ó
Ó
Ó
Ó
Ed
d
ÙE×d l
ÙE×d l
Energy stored in the dielectric
Ó
Ó
W = Q V = à à à Ρv dv à E × d l
Ó
Ó
Ó
Ó
by Gauss' Law à à D × d S = à à Ρv dv, D = Ε E
Ó
Ó Ó
Ó
W = à à D × d S à E × d l = à à à Ε E2 dv
Energy density : we = Ε E2
14
lect15new.nb
Electric fields at the boundaries of two insulators
Ó
Ó
Ó
Ó
Boundary conditions : E1 t = E2 t , D1 n - D2 n = Ρs , Ρs is usually 0 so often,
Ó
Ó
Ó
Ó
D1 n - D2 n = 0 or Ε1 E1 n = Ε2 E2 n
Ó
Ó
D1 n - D2 n = 0 Proof :
The height of this little box is Dh ® 0. And the width is Dw
A = Area
Ó
Ó
¨ D × d s = Q, therefore, D1 n A - D2 n A = Ρs A Hwith Dh = 0L, therefore
D1 n - D2 n = Ρs
Ó
Ó
E1 t = E2 t Proof :
15
lect15new.nb
Ó
Ó
¨ E × d l = 0, Electric fields are conservative
Ó
Ó
¨ E×d l =
BÓ
CÓ
DÓ
AÓ
Ó
Ó
Ó
Ó
0 = à E×d l+à E×d l+à E×d l+à E×d l =
A
B
-E1 t Dw - E1 n
C
Dh
2
- E2 n
Dh
2
D
+ E2 t Dw + E1 n
Dh
2
+ E2 n
Dh
2
= -E1 t Dw + E2 t Dw = 0,
therefore
E1 t = E2 t
Ó
Perfect conductor, E = 0, since E2 t , E1 t = 0,
therefore the E - field in a dielectric next to a perfect conductor
only has a normal component
16
lect15new.nb
Problem 5.9 Two extensive homogeneous isotropic dielectric meet on plane z =
0. For
z > 0, er1 = 4 and for z < 0,
er2 = 3. A uniform electric field E1 = 5 ax - 2 ay + 3 az kV  m exists for.Find
HaL E2 for z < 0
HbL The angles E1 and E2 make with the interface
HcL The energy densities in J  m3 in both dielectrics
HdL The energy within a cube of side 2 m centered at H3, 4, -5L
Whats E2?
E1 = 85, -2, 3<
{5, -2, 3}
E1 t = E2 t
E2 t = 85, -2, 0<
{5, -2, 0}
D1 n - D2 n = Ρs
17
lect15new.nb
Ρs = 0
D2 n = Ε2 E2 n = Ε1 E1 n
Εr1 = 4; Εr2 = 3;
E1 n = 80, 0, 3<; E2 n = Εr1 E1 n  Εr2
{0, 0, 4}
E2 = E2 t + E2 n
{5, -2, 4}
Θ1 = 90 - ArcTan@3  [email protected] 180  Pi
60.8784
Θ2 = 90 - ArcTan@4  [email protected] 180  Pi
53.3957
Θ1 = ArcTan@Sqrt@29D  3D 180.  Π
60.8784
Θ2 = ArcTan@Sqrt@29D  4D 180.  Π
53.3957
Energy density, We = 1  2 Ε
E1 = 85, -2, 3<
E
^2
18
lect15new.nb
E1 = Sqrt@25 + 4 + 9D  N
6.16441
E2
{5, -2, 4}
E2 = Sqrt@25 + 4 + 16D  N
6.7082
Ε1 = 4 ´ 8.854 ´ 10^-12; Ε2 = 3 ´ 8.854 ´ 10^-12;
we1 = 1  2 Ε1 E1^2 10^6
0.000672904
we2 = 1  2 Ε2 E2^2 10^6
0.000597645
Energy inside a cube of side 2 m centered in H3, 4, -5L
Cube is totally in region 2
volume of the cube = 8
we2 8
0.00478116
19
lect15new.nb
5.10 Region y < 0 consists of a perfect conductor while region y >
0 is a dielectric medium He1r = 2L .If there is a surface charge of 2
nC  m2 on the conductor, determine E and D at
HaL A H3, -2, 2L
HbL B H-4, 1, 5L
Region 1 is a perfect conductor; E1 = 0
y = 0 is the interface
Region 2 Εr2 = 2
Ρs = 2 ´ 10^-9 C  m^2
In region 2, D2n - D1n = Ρs; D1n = 0; D2n = 2 ´ 10^-9 C  m^2;
E2n = D2n  Ε2 = 2 ´ 10^-9  H 2 ´ 8.854 ´ 10^-12L
E2n = 2 ´ 10^-9  H 2 ´ 8.854 ´ 10^-12L 80, 1, 0<
80, 112.943, 0<
E2t = 0
At y = -2, E = 0
At y = 1, E = 80, 112.94330246216401`, 0< V  m;
D = 80, 2 ´ 10^-9, 0< C  m^2